POJ 1273 Drainage Ditches (最大流)

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题意

求一个边小于200点小于200的图的最大流。

思路

因为这里点比较少，直接用Edmonds_Karp(BFS)求就可以了，复杂度为O(E2V)，bfs的复杂度是O(E)，增广路的次数为V*E
Ford-Fulkerson(DFS)的增广路次数为C，复杂度为O(E*C)，不适用于这题。

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>using namespace std;#define LL long long#define Lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int Mod = 1000000007;const int maxn = 1e5 + 10;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;int n, m;int start, ed;int g[220][220];int path[220], flow[220];int bfs(){    queue<int> q;    memset(path, -1, sizeof(path));    path[start] = 0;    flow[start] = INF;   //源点有很多边相连    q.push(start);    while (!q.empty())    {        int t = q.front();        q.pop();        if (t == ed) break;        for (int i = 1; i <= m; i++)        {            if (i != start && path[i] == -1 && g[t][i])            {                flow[i] = min(flow[t], g[t][i]);                q.push(i);                path[i] = t;            }        }    }    if (path[ed] == -1) return -1;    return flow[m];}int Edmonds_Karp(){    int max_flow = 0, step = 0, now, pre;    while ((step = bfs()) != -1)    {        max_flow += step;        now = ed;        while (now != start)        {            pre = path[now];            g[pre][now] -= step;   //更新正向边流量            g[now][pre] += step;   //添加反向弧            now = pre;        }    }    return max_flow;}int main(){    //freopen("H:\\in.txt","r",stdin);    //freopen("H:\\out.txt","w",stdout);    while (scanf("%d%d", &n, &m) != EOF)    {        memset(g, 0, sizeof(g));        for (int i = 0; i < n; i++)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            g[u][v] += w;        }        start = 1, ed = m;        printf("%d\n", Edmonds_Karp());    }    return 0;}

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